# Killing vecotrs of Schwarzschild metric

• Vrbic
In summary, the conversation revolves around the possibility of finding Killing Vectors in a specific way from the Schwarzschild metric. The speaker suggests splitting the metric and finding KV in the spherical part, but the other person points out that this is not possible and also mentions other trivial KV that arise from spherical symmetry. The first speaker then discusses a possible transformation and its implications, but it is ultimately deemed impossible.
Vrbic
Hello,
I have a question, whether is possible to looking for Killing Vectors (KV) in this way (I know about general solution):
From Schwarzschild metric I can see two KV $\frac{\partial}{\partial t}$ and $\frac{\partial}{\partial\phi}$. Then I see that other trivial KV arent there. Metric in dt and dr is independant on $\theta, \phi$ so I suppose I can "split" metric and looking for KV just in spherical part $d\theta^2+\sin^2{\theta}d\phi^2$.
Can I suppose transformation this metric to the form: $d\alpha^2+d\beta^2$ and claim the $\frac{\partial}{\partial\alpha}, \frac{\partial}{\partial\beta}$ are KV?

No. You won't be able to find such a transformation.

Those aren't the only trivial Killing fields by the way. The generators of rotations, that is, the basis of ##so(3)##, are also easily seen to be Killing fields of the Schwarzschild metric and they arise from the spherical symmetry.

WannabeNewton said:
No. You won't be able to find such a transformation.

Those aren't the only trivial Killing fields by the way. The generators of rotations, that is, the basis of ##so(3)##, are also easily seen to be Killing fields of the Schwarzschild metric and they arise from the spherical symmetry.

Do I need know all transformation? I thought that differencial form of it could be enough. If I suppose $\alpha=\alpha(\theta,\phi), \beta=\beta(\theta,\phi)$, than I can write $d\alpha=\alpha_{\theta}+\alpha_{\phi}$, $d\beta=\beta_{\theta}+\beta_{\phi}$, where subscript means differentation. When I put it back to the $ds^2=d\alpha^2+d\beta^2$ I get 3 eq.:
$1=\alpha_{\theta}^2+\beta_{\theta}^2$
$0=\alpha_{\theta}\alpha_{\phi}+\beta_{\theta}\beta_{\phi}$
$\sin^2{\theta}=\alpha_{\phi}^2+\beta_{\phi}^2$.
From them (if from first guess $\alpha_{\theta}=\cos{\phi}$, $\beta_{\theta}=\sin{\phi}$) I get $d\alpha=\cos{\phi}d\theta-\sin{\theta}\sin{\phi}d\phi$, $d\beta=\sin{\theta}d\theta+\sin{\theta}\cos{\phi}d\phi$.
Multiplying by $g^{\mu\nu}$ I hope can get $\frac{\partial}{\partial\alpha}$, $\frac{\partial}{\partial\beta}$. What is wrong?

Vrbic said:
Can I suppose transformation this metric to the form: $d\alpha^2+d\beta^2$ and claim the $\frac{\partial}{\partial\alpha}, \frac{\partial}{\partial\beta}$ are KV?
If you could put the metric in that form, it would be FLAT. It's not flat, it's a sphere.

1 person
Vrbic said:
From them (if from first guess $\alpha_{\theta}=\cos{\phi}$, $\beta_{\theta}=\sin{\phi}$) I get $d\alpha=\cos{\phi}d\theta-\sin{\theta}\sin{\phi}d\phi$, $d\beta=\sin{\theta}d\theta+\sin{\theta}\cos{\phi}d\phi$.

This clearly doesn't reduce to the metric on the 2-sphere: simple substitution yields ##d\alpha^2 + d\beta^2 = d\theta ^2 + \sin^2 \theta d\phi^2 + (\sin \theta \cos\phi - \sin 2\phi )\sin\theta d\theta d\phi##.

You simply won't be able to put the 2-sphere metric in the form ##ds^2 = d\alpha^2 + d\beta^2## and Bill has elucidated above why this is impossible.

1 person

## 1. What are vectors in the context of the Schwarzschild metric?

Vectors in the context of the Schwarzschild metric refer to any physical quantity that has both magnitude and direction, such as force, velocity, or acceleration. In this metric, vectors are used to describe the curvature of spacetime around a non-rotating, spherically symmetric mass.

## 2. How do you kill vectors of the Schwarzschild metric?

Killing vectors of the Schwarzschild metric can be killed or eliminated by applying certain mathematical operations, such as taking derivatives and setting them equal to zero. This process is known as finding the symmetries of the metric, and it helps to simplify equations and reveal important properties of the spacetime geometry.

## 3. What is the significance of killing vectors in the Schwarzschild metric?

The significance of killing vectors in the Schwarzschild metric lies in their ability to generate symmetries within the spacetime geometry. These symmetries correspond to conservation laws, such as energy and momentum, which are essential in understanding the behavior of particles and fields in the presence of a massive object.

## 4. Are there any limitations to killing vectors in the Schwarzschild metric?

While killing vectors are powerful tools in simplifying and analyzing the Schwarzschild metric, they do have limitations. For instance, they are only applicable in vacuum solutions, where there is no matter or energy present. Additionally, they cannot be used to study the full dynamics of a system, as they only describe the spacetime geometry and not the particles or fields within it.

## 5. How do killing vectors of the Schwarzschild metric relate to general relativity?

Killing vectors of the Schwarzschild metric play a crucial role in the theory of general relativity, as they help to describe the symmetries and geometries of spacetime. They are used to solve the Einstein field equations, which describe the relationship between the curvature of spacetime and the distribution of matter and energy within it. Therefore, understanding killing vectors is essential in understanding the fundamental principles of general relativity.

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